We are asked to simplify expressions involving square roots of negative numbers and express the results in terms of $i$. Specifically, we have the following problems: a) $\sqrt{-4} \cdot \sqrt{-16}$ b) $\sqrt{-25} \cdot \sqrt{-9}$ c) $\frac{\sqrt{-36}}{\sqrt{-4}}$ d) $\frac{\sqrt{-64}}{\sqrt{-16}}$ f) $\frac{\sqrt{-18}}{\sqrt{-3}}$

1. Problem Description

We are asked to simplify expressions involving square roots of negative numbers and express the results in terms of $i$. Specifically, we have the following problems:

a) $\sqrt{-4} \cdot \sqrt{-16}$

b) $\sqrt{-25} \cdot \sqrt{-9}$

c) $\frac{\sqrt{-36}}{\sqrt{-4}}$

d) $\frac{\sqrt{-64}}{\sqrt{-16}}$

f) $\frac{\sqrt{-18}}{\sqrt{-3}}$

2. Solution Steps

Recall that $i = \sqrt{-1}$ and $i^2 = -1$.

a) $\sqrt{-4} \cdot \sqrt{-16} = \sqrt{4}i \cdot \sqrt{16}i = 2i \cdot 4i = 8i^2 = 8(-1) = -8$

b) $\sqrt{-25} \cdot \sqrt{-9} = \sqrt{25}i \cdot \sqrt{9}i = 5i \cdot 3i = 15i^2 = 15(-1) = -15$

c) $\frac{\sqrt{-36}}{\sqrt{-4}} = \frac{\sqrt{36}i}{\sqrt{4}i} = \frac{6i}{2i} = \frac{6}{2} = 3$

d) $\frac{\sqrt{-64}}{\sqrt{-16}} = \frac{\sqrt{64}i}{\sqrt{16}i} = \frac{8i}{4i} = \frac{8}{4} = 2$

f) $\frac{\sqrt{-18}}{\sqrt{-3}} = \frac{\sqrt{18}i}{\sqrt{3}i} = \frac{\sqrt{18}}{\sqrt{3}} = \sqrt{\frac{18}{3}} = \sqrt{6}$

3. Final Answer

a) -8

b) -15

c) 3

d) 2

f) $\sqrt{6}$